The Electromagnetic Radiation


Problems and Solutions

Wien's Law

(1) The sun has a temperature of 5800K. What is the sun's wavelength of maximum emission? Solution

(2) The Earth has an average temperature of 288K. What is the Earth's wavelength of maximum emission? Solution

Stefan Boltzmann Law

(3) How much more energy is emitted by an object that has a temperature of 100K compared to an object that has a temperature of 10K? Solution

Electromagnetic Wave Problems

After learning about electromagnetic waves and its properties, we will now test you on your knowledge. You were introduced to two very popular equations concerning electromagnetic energy: the speed of light and the energy emitted from photons. Using the equations below, you will solve basic electromagnetic problems:
C = f * λ .and. Δ E = h * f

(4) A ray, emitted from the sun, is shining through your kitchen window into a prism. The prism then casts a rainbow on the windowsill. You just so happen to have a hand-held radiometer handy; closing your eyes you place the instrument on a specific color of the rainbow. You then open your eyes to see that the radiometer measured the energy from that color at 3.0 x 10-19 joules. Given Plank's constant of 6.6256 x 10-34 joules/sec, what possible color did the radiometer measure? How do we know this?

Hints: 1/sec or sec-1 is another way of writing cycles/sec. Likewise, it is known as Hz, a unit of frequency. One Hz is equal to one cycle/sec, or one sec-1. The visible spectrum is measured from 4.3 x 1014 Hz (red colors) to 7.5 x 1014 Hz (violet colors); so you have a bit of lead-way with your possible color. Solution

As stated before, light is measured in photons. These photons are measured in small units of length called nanometers (nm). Let's combine both equations and really test your comprehension.

(5) Your teacher dropped a bag of Skittles on the floor: what a mess! She thought, "What a perfect problem for my intelligent students!" Picking up one of the tasty candies, she carefully measures the wavelength of one of the colors at 600 nm: boy is she ever smart! Her question is: how many joules of energy are contained in a photon with a wavelength of 600 nm? Solution


Wien's Law

(1) 0.5 microns

Wavelength of maximum emission = 2900 / Object's temperature in Kelvin
Sun's wavelength of maximum emission = 2900 / 5800 = 0.5 microns

(2) 10 microns

Wavelength of maximum emission = 2900 / Object's temperature in Kelvin
Earth's wavelength of maximum emission = 2900 / 288 = 10 microns

Stefan Boltzmann Law

(3) 10,000 times more energy

The amount of energy emitted by an object is directly related to the fourth power of the object's temperature. E = T^4, where E is the amount of energy emitted by an object per square meter per second and T is the temperature of the object in Kelvin.

E = T4

E = (100)4 = (102)4 = 108 for the object with a temperature of 100K

E= (10)4 = 104 for the object with a temperature of 10K

Energy emitted by object with temperature of 100K / Energy emitted by object with temperature of
10K = 108/104 = 104 = 10,000 times more energy emitted by the object that has a temperature of 100K.

Electromagnetic Wave Problems

(4) Solution in detail below:

First we need to consider what would be the best equation to use. Obviously, the equation involving energy change, Plank's constant, and frequency is the best way to go. Next, we need to figure out what we are solving for. In this problem, they ask for the possible color that you measured. If we relate the energy with Plank's constant, we can solve for frequency. We are given the energy, 3.0 x 10-19 joules, as well as Plank's constant, 6.6256 x 10-34 joules/sec. Also, we are given the frequencies emitted by the visible spectrum, from red to violet. This problem is too easy now! If we solve for the frequency, we can then relate it to the energy emitted, measured in either sec-1, Hz, or less commonly, cycles/sec. Let's solve:

- What is the best equation to use? λ E = h* ƒ

- Solve for the intended variable. λ E/ h = ƒ

- What is given? Energy change = 3.0 x 10-19 joules
Plank's constant = 6.6256 x 10-34 joules/sec
Visible spectrum frequencies

- Do the math.
(3.0 x 10-19 joules) / 6.6256 x 10-34 joules/sec = ƒ
- Joules cancel out with joules, and one is left with sec-1, a frequency.
Answer = 4.5 x 1014 sec-1

- Answer the problem:
If the math is done correctly one should get 4.5 x 1014 sec1. This falls within the given visible spectrum frequencies. Being a low frequency on the visible spectrum, close to the frequencies emitted by red colors, one could assume that you measured a color close to red; though orange and yellow cannot be out of the question as the given frequencies are a bit vague.

(5) 3.3128 x 10-19 joules

Again there are many conversions in this problem. Let's attack it piece by piece. First we have an equation, the speed of light, 3.0 x 1010 cm/sec, and the given wavelength, 600 nm. We can relate these two parts once they have units that can cancel. When they have similar units, we can solve for one of the equations, C = ƒ * λ, and then use that information to solve for the other equation, Δ E = h * ƒ. In the first equation, we need to solve for frequency, and then take that number and plug it into the second equation, solving for Δ E. Let's solve:

- What are we solving for in the end? Δ E

- What is given? C = ƒ * λ
Δ E = h * ƒ
C= 3.0 x 1010 cm/sec
λ = 600 nm
h = 6.6256 x 10-34 joules/sec

- Do the math

3.0 x 1010 cm/sec * 1 m/100cm = 3.0 x 108 m/sec
600 nm * 1 x 10-9 m/nm = 6.00 x 10-7 m
C / λ = ƒ
(3.0 x 108 m/sec) / 6.00 x 10-7m
ƒ = 5.0 x 1014 sec-1

Δ E = h * ƒ
6.6256 x 10-34 joules/sec * 5.0 x 1014 sec-1
Answer = 3.3128 x 10-19 joules